Problem: The roots of
\[z^7 = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\]are $\text{cis } \theta_1$, $\text{cis } \theta_2$, $\dots$, $\text{cis } \theta_7$, where $0^\circ \le \theta_k < 360^\circ$ for all $1 \le k \le 7$.  Find $\theta_1 + \theta_2 + \dots + \theta_7$.  Give your answer in degrees.
Let $z = \text{cis } \theta$.  Then
\[z^7 = \text{cis } 7 \theta.\]Using cis notation,
\[-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} = \text{cis } 225^\circ,\]so we want
\[\text{cis } 7 \theta = \text{cis } 225^\circ.\]This equation holds if and only if
\[7 \theta = 225^\circ + 360^\circ k\]for some integer $k$, or
\[\theta = \frac{225^\circ + 360^\circ k}{7}.\][asy]
unitsize(2 cm);

draw((-1.2,0)--(1.2,0));
draw((0,-1.2)--(0,1.2));
draw(Circle((0,0),1));

dot("cis $\frac{225^\circ}{7}$", dir(225/7), dir(225/7));

for(int i = 1; i <= 6; ++i) {
  dot(dir(225/7 + 360*i/7));
}

label("Re", (1.2,0), NE);
label("Im", (0,1.2), NE);
[/asy]

The angles of this form that are between $0^\circ$ and $360^\circ$ are
\[\frac{225^\circ}{7}, \quad \frac{225^\circ + 360^\circ}{7}, \quad \frac{225^\circ + 2 \cdot 360^\circ}{7}, \quad \dots, \quad \frac{225^\circ + 6 \cdot 360^\circ}{7}.\]By the formula for an arithmetic series, the sum of these angles is
\[\frac{1}{2} \cdot \left( \frac{225^\circ}{7} + \frac{225^\circ + 6 \cdot 360^\circ}{7} \right) \cdot 7 = \boxed{1305^\circ}.\]